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A coin is tossed 7 times. The number of times head comes is 5, so the probability of getting tail is
(a.) $ \frac{2}{7} $
(b.) $ \frac{5}{7} $
(c.) $ \frac{1}{7} $
(d.) $ \frac{4}{7} $
Given:
A coin is tossed 7 times. The number of times head comes is 5.
To do:
We have to find the probability of getting a tail.
Solution:
The total number of outcomes$=7$.
Number of outcomes in which head comes$=5$
This implies,
Number of outcomes in which tail comes$=7-5=2$
We know that,
Probability of an event=$ \frac{Number \ of \ favourable \ outcomes}{Total \ number \ of \ outcomes}$
Therefore,
Probability of getting tail$=\frac{2}{7}$.
The probability of getting tail is $\frac{2}{7}$.
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